What is the general solution for sin(2t)cos(t)=0,0<= t<= pi ?

The general solution for sin(2t)cos(t)=0,0<= t<= pi is t=0,t= pi/2 ,t=pi

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sin(2t)cos(t)=0,0<= t<= pi

sin (2 t) cos (t) = 0, 0 \leq t \leq π

t = 0, t = \frac{π}{2}, t = π

Solution steps

sin (2 t) cos (t) = 0, 0 \leq t \leq π

Solving each part separately

sin (2 t) = 0 or cos (t) = 0

sin (2 t) = 0, 0 \leq t \leq π : t = 0, t = \frac{π}{2}, t = π

cos (t) = 0, 0 \leq t \leq π : t = \frac{π}{2}

Combine all the solutions

t = 0, t = \frac{π}{2}, t = π

2 sin^{3} (x) = sin^{3} (x)

θ, y = 4 sin (θ)

- 3 (sin^{2} (θ) - cos^{2} (θ)) = 0

cos^{2} (t) - sin^{2} (t) = - 1

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

What is the general solution for sin(2t)cos(t)=0,0<= t<= pi ?
The general solution for sin(2t)cos(t)=0,0<= t<= pi is t=0,t= pi/2 ,t=pi